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To calculate the 90% confidence interval for the population mean weight of bags of potato chips, we'll use the formula for a confidence interval:

\[
\text{Confidence Interval} = \bar{x} \pm z_{\alpha/2} \left( \frac{s}{\sqrt{n}} \right)
\]

Where:
- \(\bar{x} = 7.94\) ounces (sample mean),
- \(s = 0.013\) ounces (sample standard deviation),
- \(n = 12\) (sample size),
- \(z_{\alpha/2} = 1.645\) for a 90% confidence interval (since it's a two-tailed test and based on the normal distribution).

### Step-by-Step:
1. **Calculate the standard error (SE):**
   \[
   SE = \frac{s}{\sqrt{n}} = \frac{0.013}{\sqrt{12}} = \frac{0.013}{3.464} \approx 0.00375
   \]

2. **Determine the margin of error (MOE):**
   \[
   MOE = z_{\alpha/2} \times SE = 1.645 \times 0.00375 \approx 0.00617
   \]

3. **Find the confidence interval:**
   - Lower limit: \(7.94 - 0.00617 \approx 7.93\)
   - Upper limit: \(7.94 + 0.00617 \approx 7.95\)

Thus, the 90% confidence interval is approximately **(7.93, 7.95)** ounces.

Would you like any further clarification or additional details on this calculation?

### Follow-up questions:
1. What is the significance of a confidence interval in hypothesis testing?
2. How would the confidence interval change with a 95% confidence level?
3. Can you explain the role of the sample size in determining the width of a confidence interval?
4. Why do we assume the distribution of weights is approximately normal in this scenario?
5. How does the z-value change for different confidence levels?

**Tip:** The confidence interval gives a range of values within which the true population mean is likely to fall with a certain level of confidence (here, 90%).

A potato chip company wants to evaluate the accuracy of its potato chip bag-filling machine. Bags are labeled as containing 8 ounces of potato chips. A simple random sample of 12 bags had mean weight 7.94 ounces with a sample standard deviation of 0.013 ounce. Assume the weights are approximately normally distributed. Construct a 90% confidence interval for the population mean weight of bags of potato chips. Round the answers to at least two decimal places.

Calculate the 90% confidence interval for the mean weight of potato chip bags using sample data of 12 bags. The sample mean is 7.94 ounces with a standard deviation of 0.013 ounces. This problem involves key concepts in statistics, including the confidence interval formula and normal distribution assumptions.

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